Question 1117095
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You have a triangle with a segment from one vertex to the opposite side; and you know -- or have enough information to determine -- the lengths of all but one of the segments in the figure.  In any problem like that, the use of Stewart's Theorem is a possible path to the solution.<br>
Let the triangle be ABC, and let AD be a segment with D on BC.  Then Stewart's Theorem says<br>
{{{AB*CD*AB+AC*BD*AC=AD*BC*AD+BD*DC*BC}}}<br>
Note the theorem holds for any such segment AD -- it can be a median, or an angle bisector, or an altitude; but it can also be ANY segment AD with D on BC.<br>
In your problem, AD is the angle bisector.  An angle bisector in a triangle divides the opposite side into two parts whose lengths are in the same ratio as the lengths of the two sides that form the angle.<br>
With AB = 50, AC = 60, and BC = 70, the angle bisector AD divides BC with length 70 into two pieces whose lengths are in the ratio 50:60, or 5:6.  That makes the lengths of the two segments BD = 350/11 and CD = 420/11.<br>
Then you are ready to plug in the segment lengths into the formula in Stewart's Theorem:
AB=50; AC=60; BC=70; BD=350/11; CD=420/11;
the unknown x is the length of AD<br>
Plugging in those values leads to the answer of 42.25 that you show for the length of AD.  I leave it to you to do the calculations.